L´od´z University Press 2013, 51 – 55
ON SMOOTH HYPERSURFACES CONTAINING A GIVEN SUBVARIETY
ZBIGNIEW JELONEK
Abstract. We reprove some results about affine complete intersections.
1. Introduction.
Let k be an algebraically closed field. Let Xn be a smooth affine variety (of
dimension n). Let us recall that a variety H ⊂ X is a hypersurface if the ideal I(H) ⊂ k[X] is generated by a single polynomial. Let Yr ⊂ Xn be a smooth
subvariety. It was proved in [2] (see also [3]), that if n ≥ 2r + 1 then there is a smooth complete intersection Z2r⊂ Xn such that Yr⊂ Z2r. In general this result
can not be improved- see Example 2.2. We also show how to use results from [6] to improve the result above in some special cases. In particular we show:
Theorem 1.1. (Greco, Valabrega) Let Xn be a smooth variety and let Yr be a
smooth subvariety of X. Assume that the rth Chow group CHr(Yr) vanishes. If
n ≥ 2r, then there is a smooth complete intersection Z2r−1 ⊂ X such that Yr ⊂
Z2r−1.
and
Theorem 1.2. (Murthy) Let Yr ⊂ An
be a smooth subvariety. If n ≥ 2r then there is a smooth hypersurface H ⊂ An such that Y ⊂ H.
In particular a smooth surface S ⊂ A4 is contained in a smooth hypersurface H ⊂ A4. Let us note that this is not true in the projective case: it is well known that a smooth surface S ⊂ P4 is not contained in any smooth hypersurface H ⊂ P4,
2010 Mathematics Subject Classification. 14R10, 14R99.
Key words and phrases. Algebraic vector bundle, complete intersection, unimodular row. The author was partially supported by the grant of Polish Ministry of Science 2010-2013.
51
unless it is a complete intersection. Our approach is slightly different than the original ones.
2. Main Result. We start with:
Theorem 2.1. Let Y ⊂ X be smooth affine varieties. Then there is a smooth hypersurface V (f ) ⊂ X which contains Y if and only if the normal bundle of Y contains a one dimensional trivial summand i.e.,
NX/Y = T ⊕ E1,
where E1 denotes a trivial line bundle.
Proof. Assume that there is a smooth hypersurface H = V (f ) ⊂ X which contains Y. We have
T Y ⊂ T H ⊂ T X, in particular
NX/Y = NH/Z⊕ NX/H|Y.
However, the normal bundle of the smooth hypersurface H = V (f ) is trivial (in fact the class of f is a generator of the conormal bundle of H).
Conversely, assume that
NX/Y = T ⊕ E1.
Hence also
N∗X/Y = T∗⊕ E1.
This means that the conormal bundle N∗X/Y has a nowhere vanishing section s ∈ Γ(Y, N∗X/Y). But Γ(Y, N∗X/Y) = I(Y )/I(Y )2, where I(Y ) ⊂ k[X] denotes the ideal of the subvariety Y. Hence s determines a polynomial s ∈ I(X) such that the class of s is s. Take a point a ∈ Y and local coordinates (u1, ..., un) at a such that Y is
described by local equations u1, ..., ut (t = codimY ) near a. Since u1, ..., ut freely
generate the bundle N∗X/Y near the point a, we have
s =
t
X
i=1
αiui,
where αi∈ k[Ua] (Uadenotes some open neighborhood of a in Y ). Since the section
s nowhere vanishes, there exists at least one i0 such that αi0 6= 0. Let us compute
the derivative dys of the polynomial s at the point y ∈ Y. We have
s =
t
X
i=1
hence there are polynomials fj, hj∈ I(Y ), j = 1, ..., m, such that s = t X i=1 αiui+ m X j=1 fjhj.
Now we easily see that
das = t
X
i=1
αidaui.
Since daui, i = 1, ..., n, are linearly independent and not all αi vanish at y we have
dys 6= 0. Hence the hypersurface V (s) is smooth along Y. Let I(Y ) = (g1, ..., gr).
Consider the linear system on X given by the polynomials (s, g2
1, ..., g2r). The base
locus of this system is exactly the subvariety Y. We can extend the set {g2 1, ..., gr2}
adding some polynomials {g2
jαi, j = 1, ..., s, i = 0, 1, ..., k} in such a way that a
new system (s, g21, ..., g2r, g2jαi) is unramified outside Y. Indeed, let x ∈ X \ Y. There
is a polynomial gx∈ I(Y ), such that gx(x) 6= 0. Let α1, ..., α2k+1 (k = dim X) be
polynomials which gives an embedding of X into k2n+1. In some neighbourhood
Ux of X we still have gx 6= 0. Since X \ Y is quasi-compact we can cover X \ Y
by a finite set Uxi, i ∈ I of such neighbourhoods. Associate with every such Ux
the set Sx := {g2x, gx2α1, ..., g2xα2k+1}. It is easy to see, that the system given by
polynomials {s, g2
1, ..., g2r} ∪
S
i∈ISxi is unramified on X \ Y.
Hence by the Bertini Theorem (see [4], Corollary 12 and [5], Theorem 3.1) the hypersurface V (s +Pr
i=1βig2i +P βj,sgj2αs) for generic βi, βj,s is smooth outside
Y. But for y ∈ Y, dy(s + r X i=1 βig2i + X βj,sg2jαs) = dys.
This implies that the hypersurface V (s +Pr
i=1βig2i +P βj,sgj2αs) is also smooth
along Y. Hence we can take f = s +Pr
i=1βig2i +P βj,sgj2αs.
Let X2n be a smooth variety and Yn be a smooth subvariety of X2n. We show that in general does not exist a smooth hypersurface H ⊂ X2n, such that Yn⊂ H. Indeed we have:
Example 2.2. Let Hd ⊂ Pn+1 be a smooth hypersurface of degree d > n + 2.
Let Y ⊂ H be an affine open subset. By [7] we have CHn(Y ) 6= 0. Take a
non-zero z ∈ CHn(Y ). By Riemann-Roch without denominators and Serre Splitting
Theorem ( Theorem 2.3 below), there exists an algebraic vector bundle F on Y of rank n such that cn(F) = (n − 1)!z. Since CHn(Y ) has no (n − 1)! torsion (see e.g.
[6]) we have cn(F) 6= 0. Now let X denote the total space of this vector bundle.
Then Y ⊂ X (as the zero-section) and NX/Y ∼= F. Since the top Chern class of F
does not vanish, the bundle F does not have a one dimensional trivial summand. In particular Y is not contained in any smooth hypersurface in X (see Theorem 2.1).
In the sequel we need the following ( see [1], p.177, Th. 7.1.8 and [5], Corollary 3.4):
Theorem 2.3. (Serre Splitting Theorem) Let X be a smooth affine variety and let F be an algebraic vector bundle on X. If rank F > dim X, then F has a one dimensional trivial summand i.e.,
F = T ⊕ E1. Now we are in a position to prove:
Theorem 2.4. Let Xn be a smooth variety and let Yr be a smooth subvariety of
X. If n ≥ 2r + 1 then there is a smooth complete intersection Z2r ⊂ Xn such
that Yr ⊂ Z2r. Assume additionally that the rth Chow group CHr(Yr) vanishes.
If n ≥ 2r, then there is a smooth complete intersection Z2r−1 ⊂ X such that
Yr⊂ Z2r−1.
Proof. Assume first that s = n − 2r > 0. Since dim Yr< rank NX/Y, Theorem 2.3
shows that NX/Y = T ⊕ E1, where E1 denotes a trivial line bundle. By Theorem
2.1 there exists a smooth hypersurface H = V (f ) (where f is a reduced polynomial) such that Y ⊂ H. Now we can apply the mathematical induction. This completes the proof of the first part of Theorem 2.4.
For the proof of the second part let us note that the bundle F = N∗Z2r/Yr has
a one dimensional trivial summand as cr(F) = 0, by the Theorem of Murthy (see
[6], Th. 3.8). Now we can finish by applying Theorem 2.1.
Theorem 2.5. Let Xn, Yrbe as above. If n ≥ 2r + 1 then there is a smooth hyper-surface H = V (f ) such that Yr ⊂ H. If the rth Chow group CHr(Xn) vanishes,
then it is enough to assume n ≥ 2r.
Proof. It is enough to consider only the last statement. Moreover, we can assume that n = 2r. Let Yr=Ss
i=1Yi be the decomposition of Y into irreducible
compo-nents. Of course Yi∩ Yj = ∅ for i 6= j. We show that the bundle F = NX/Y has
a one dimensional trivial summand over every Yi. Indeed, if dim Yi < r then it
follows from the Serre Splitting Theorem. Assume that dim Yi= r. Let ι : Yi→ X
be the inclusion. By the self-intersection formula we have the following expression for the top Chern class of the normal bundle of Yi:
cr(F|Yi) = ι
∗◦ ι ∗[Yi],
where [Yi] ∈ CH0[Yi] = Z is a generator. By our assumption we have cr(F|Yi) = 0.
Now by the Theorem of Murthy, invoked above, the normal bundle NX/Y splits
over Yi in a suitable way. Finally we can use Theorem 2.1.
The last statement of Theorem 2.5 can be applied to X = An, or more generally to X = open affine subset of An. In particular we have:
Corollary 2.6. Let Yr ⊂ An
be a smooth subvariety. If n ≥ 2r then there is a smooth hypersurface H ⊂ An such that Y ⊂ H.
Theorems above suggest that if all (positive) Chow groups of X and Y vanish, then it is easier to find a smooth hypersurface which contains a given smooth subvariety Y ⊂ X. However, we show that also in that case there are examples of smooth subvarieties Y ⊂ X which are not contained in any smooth hypersurface of X. In our example X will be an open affine subset of A9 and Y be an affine open subset of A7. In particular Y and X have all positive Chow groups trivial. Example 2.7. Consider the variety Γ = {(x, y) ∈ k3× k3 : P3
i=1xiyi = 1}. By
the Raynaud Theorem (see [8] and [9]) the algebraic vector bundle given by the unimodular row (x1, x2, x3) is not free. Let Λ = {(x, y) ∈ k3× k3:P3i=1xiyi= 0}
be an affine cone and let Y0 = A6\ Λ. Hence Y0 is an affine open subset of A6. Moreover, the algebraic vector bundle F given by the unimodular row (x1, x2, x3) is
not trivial, because it is not trivial after restriction to Γ. Since every stably trivial line bundle is trivial and rank F = 2, we see that the vector bundle F does not split.
Take Y00= Y0× k, X = Y0× k3 and consider the embedding
φ : Y003 ((x, y), t) 7→ ((x, y), x1t, x2t, x3t) ∈ X.
Take Y = φ(Y00). By direct computations we see that the normal bundle NX/Y
restricted to the subvariety Y0× {0} is equal to E3/ < (x1, x2, x3) >∼= F
(where Esdenotes the trivial bundle of rank s). Since the bundle F does not split, neither does NX/Y. In particular Y is not contained in any smooth hypersurface
in X. Moreover, X is an open subset of A9 and Y is isomorphic to an open subset of A7.
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Instytut Matematyczny PAN, ul. ´Sniadeckich 8, 00-956 Warszawa, Poland, E-mail address: najelone@cyf-kr.edu.pl
